Daily Challenge #133 - Suitcase Packing

Mr. Square is going on holiday. He wants to bring 2 of his favorite squares with him, so he put them in his rectangle suitcase.

Write a function that, given the size of the squares and the suitcase, return whether the squares can fit inside the suitcase.

fit_in(a,b,m,n)
a,b are the sizes of the squares
m,n are the sizes of the suitcase

Example
fit_in(1,2,3,2) should return True
fit_in(1,2,2,1) should return False
fit_in(3,2,3,2) should return False
fit_in(1,2,1,2) should return False

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Comments

[Andreas Jakof]

The long side is the obvious case, but what if

• Square A: 1x1
• Square B: 2x2
• Rectangle 3x1

=> FitIn(1,2,3,1)?

in C#

public bool FitIn(int a, int b, int m, int n)
{
   //both squares (added) must fit into the long side of the rectangle AND
   //the bigger single square must fit into the short side of the rectangle

   int longSide = Math.Max(m,n);
   int shortSide = Math.Min(m,n);
   int addedSquareLengths = a + b;
   int biggerSingleSquareSize = Math.Max(a,b);

   //do they fit?
   bool fitsLong = addedSquareLengths <= longSide;
   bool fitsShort = biggerSingleSquareSize <= shortSide;

   return fitsLong && fitsShort;
}

I know, it's a bit verbose, but very self explanatory. A short version will look like this:

public bool FitIn(int a, int b, int m, int n) =>
   (a + b <= Math.Max(m,n)) && //longSide
   (Math.Max(a,b) <= Math.Min(m,n)); //shortSide

[Vaibhav Yadav]

In JavaScript.

const fit_in = (a, b, m, n) => a + b <= Math.max(m, n);

[Andreas Jakof]

Assert.IsFalse(fit_in(1,2,3,1)

[Nick Holmes]

In F#.

let fit_in a b m n = a + b <= max m n

[Andreas Jakof]

let actual = fit_in 1 2 3 1
Assert.That(actual, Is.False)

[erezwanderman]

JS

fit_in = (a,b,m,n) => ((a + b) <= m && a <= n && b <= n) || ((a + b) <= n && a <= m && b <= m)

[Vaibhav Yadav]

Thanks for pointing that out, edited the solution.